Convexity, concavity, super-additivity, and sub-additivity of cost function without fixed cost

Transcription

1 MPRA Munich Personal RePEc Archive Convexity, concavity, super-additivity, and sub-additivity of cost function without fixed cost Yasuhito Tanaka and Masahiko Hattori 31 July 2017 Online at MPRA Paper No , posted 1 August :30 UTC

2 Convexity, concavity, super-additivity, and sub-additivity of cost function without fixed cost Masahiko Hattori* Faculty of Economics, Doshisha University, Kamigyo-ku, Kyoto, , Japan. and Yasuhito Tanaka Faculty of Economics, Doshisha University, Kamigyo-ku, Kyoto, , Japan. Abstract With zero fixed cost, convexity of a cost function implies super-additivity, and concavity of a cost function implies sub-additivity. But converse relations do not hold. However, in addition to the zero fixed cost condition we put the following assumption. (1) If a cost function is convex in some interval, it is convex throughout the domain. (2) If a cost function is concave in some interval, it is concave throughout the domain. Then, super-additivity implies convexity and sub-additivity implies concavity. Subsequently, super-additivity and convexity are equivalent, and sub-additivity and concavity are equivalent. Keywords: cost function, convexity, concavity, super-additivity, sub-additivity JEL Classification: D43, L13 *mhattori@mail.doshisha.ac.jp yasuhito@mail.doshisha.ac.jp

3 1 Introduction Convexity and concavity are important properties for cost functions of firms. Also superadditivity and sub-additivity are other important properties for them. A cost function c(x) is convex when it satisfies c(λx + (1 λ)y) λc(x) + (1 λ)c(y) for 0 λ 1, x 0, y 0. It is concave when it satisfies c(λx + (1 λ)y) λc(x) + (1 λ)c(y) for 0 λ 1, x 0, y 0. It is super-additive if it satisfies It is sub-additive if it satisfies c(x + y) c(x) + c(y), for x 0, y 0. c(x + y) c(x) + c(y), for x 0, y 0. It is well known that with zero fixed cost, that is, c(0) = 0, convexity implies superadditivity, and concavity implies sub-additivity. But converse relations do not hold. See Bruckner and Ostrow (1962) and Sen and Stamatopoulos (2016). Referring to Bourin and Hiai (2015), Sen and Stamatopoulos (2016) pointed out that the following function is super-additive but it is not convex. xe 1 x 2, x 0. However, if, in addition to the zero fixed cost condition, we put the following assumption, we can show that super-additivity implies convexity, and sub-additivity implies concavity. Assumption 1. the domain. (1) If a cost function is convex in some interval, it is convex throughout (2) If a cost function is concave in some interval, it is concave throughout the domain. Then, super-additivity and convexity are equivalent, and sub-additivity and concavity are equivalent. Assumption 1 excludes a case where a cost function is strictly convex in some interval is such a function. We assume that a cost function, c(x), is positive for positive x, increasing and continuous. and strictly concave in another interval. Above mentioned xe 1 x 2 2

4 2 Main results We show the following theorems. Theorem 1. (1) If there is no fixed cost, convexity of a cost function implies its superadditivity. (2) If there is no fixed cost, concavity of a cost function implies its sub-additivity. Proof. (1) Convexity super-additivity: For 0 λ 1 and x 0 convexity implies c(λx + (1 λ) 0) λc(x) + (1 λ)c(0) = λc(x). Thus, Then, for x > 0 and y 0 c(λx) λc(x). (2) Concavity sub-additivity: c(x) + c(y) =c ( x (x + y)) + c ( y (x + y)) x + y x + y x c(x + y) + y c(x + y) = c(x + y). x + y x + y For 0 λ 1 and x 0 concavity implies c(λx + (1 λ) 0) λc(x) + (1 λ)c(0) = λc(x). Thus, Then, for x > 0 and y 0 c(λx) λc(x). c(x) + c(y) =c ( x (x + y)) + c ( y (x + y)) x + y x + y x c(x + y) + y c(x + y) = c(x + y). x + y x + y Theorem 2. Suppose that a cost function satisfies Assumption 1, and there is no fixed cost. (1) Super-additivity of a cost function implies convexity. (2) Sub-additivity of a cost function implies concavity. 3

5 Proof. (1) Super-additivity convexity: Let x > 0. By super-additivity, for any 0 λ 1, c(x) c(λx) + c((1 λ)x). This implies, for any 0 λ 1, c(x) c(λx + (1 λ) 0) + c(λ 0 + (1 λ)x). (1) By the zero fixed cost condition, for any 0 λ 1, c(x) = λc(x) + (1 λ)c(0) + λc(0) + (1 λ)c(x). (2) (1) and (2) imply that, for any 0 λ 1, λc(x) + (1 λ)c(0) c(λx + (1 λ) 0), (3a) or λc(0) + (1 λ)c(x) c(λ 0 + (1 λ)x) (3b) holds. Suppose that for some x and some 0 < λ < 1, (3a) does not hold, that is, λc(x) + (1 λ)c(0) < c(λx + (1 λ) 0). (4a) Then, (3b) must hold with strict inequality, that is, λc(0) + (1 λ)c(x) > c(λ 0 + (1 λ)x). (4b) (4a) means that c(x) can not be convex (nor linear) throughout [0, x]. Similarly, (4b) means that c(x) can not be concave (nor linear) throughout [0, x]. It contradicts to Assumption 1. If c(x) is not differentiable at, for example, y, it may be not concave nor convex (nor linear) in an interval including y. However, if we exclude a pathological function which is continuous everywhere but differentiable nowhere, any function is concave or convex in some interval. Then, by Assumption 1 it is concave or convex throughout the domain. Therefore, both (3a) and (3b) must hold for any x > 0 and any 0 λ 1. Alternatively, we can prove the same result assuming that (3b) instead of (3a) does not hold. Let y = αx for x > 0 and 0 < α < 1. Assume that for some 0 < λ < 1, λc(x) + (1 λ)c(y) < c(λx + (1 λ)y). (5) Let z = λx + (1 λ)y = [λ + (1 λ)α]x. 4

6 By (3a) Then, from (5) αc(x) c(αx) = c(y). ( λ + 1 λ) c(y) < c(z). α This means where β = α λ + (1 λ)α c(y) < βc(z) + (1 β)c(0), (6) > 0 and 1 β = λ(1 α) λ + (1 λ)α > 0. (5) means that c(x) can not be convex (nor linear) throughout [y, x]. (6) means that c(x) can not be concave (nor linear) throughout [0, z]. It contradicts to Assumption 1. Therefore, for any x > y 0 and any 0 λ 1 λc(x) + (1 λ)c(y) c(λx + (1 λ)y). Thus, c(x) is convex throughout the domain. (2) Sub-additivity concavity: Let x > 0. By sub-additivity, for any 0 λ 1, This implies, for any 0 λ 1, c(x) c(λx) + c((1 λ)x). c(x) c(λx + (1 λ) 0) + c(λ 0 + (1 λ)x). (7) By the zero fixed cost condition, for any 0 λ 1, c(x) = λc(x) + (1 λ)c(0) + λc(0) + (1 λ)c(x). (8) (7) and (8) imply that, for any 0 λ 1, λc(x) + (1 λ)c(0) c(λx + (1 λ) 0), (9a) or λc(0) + (1 λ)c(x) c(λ 0 + (1 λ)x) (9b) holds. Suppose that for some x and some 0 < λ < 1, (9a) does not hold, that is, λc(x) + (1 λ)c(0) > c(λx + (1 λ) 0). (10a) Then, (9b) must hold with strict inequality, that is, λc(0) + (1 λ)c(x) < c(λ 0 + (1 λ)x). (10b) 5

7 (10a) means that c(x) can not be concave throughout [0, x]. Similarly, (10b) means that c(x) can not be convex throughout [0, x]. It contradicts to Assumption 1. Therefore, both (9a) and (9b) must hold for any x > 0 and any 0 λ 1. Alternatively, we can prove the same result assuming that (9b) instead of (9a) does not hold. Let y = α x for x > 0 and 0 < α < 1. Assume that for some 0 < λ < 1, λc(x) + (1 λ)c(y ) > c(λx + (1 λ)y ). (11) Let By (9a) Then, from (11) z = λx + (1 λ)y = [λ + (1 λ)α ]x. α c(x) c(α x) = c(y ). ( λ α + 1 λ) c(y ) > c(z ). This means where β = c(y ) > β c(z ) + (1 β )c(0), (12) α λ + (1 λ)α > 0, and 1 β = λ(1 α ) λ + (1 λ)α > 0. (11) means that c(x) can not be concave throughout [y, x]. (12) means that c(x) can not be convex throughout [0, z ]. It contradicts to Assumption 1. Therefore, for any x > y 0 and any 0 λ 1 λc(x) + (1 λ)c(y ) c(λx + (1 λ)y ). Thus, c(x) is concave throughout the domain. If c(x) is twice continuously differentiable, convexity of c(x) is equivalent to non-negativity of its second order derivative, that is, c (x) 0, and its concavity is equivalent to nonpositivity of its second order derivative, that is, c (x) 0. We present a proof of Theorem 2 using differentiability. Proof of Theorem 2 with differentiability. Let x > 0 and y > 0. By super-additivity (1) Super-additivity convexity: c(x + y) c(x) + c(y). With the zero fixed cost condition, we have c(x + y) c(x) y c(y) c(0). y 6

8 Let y 0, Thus, This means Let x 0, we obtain c(x + y) c(x) lim y 0 y c (x) c (0). c (x) c (0) x c(y) c(0) lim. y 0 y 0. c (0) = lim x 0 c (x) c (0) x Suppose that there exist x 2 > x 1 > 0 such that c (x 2 ) < c (x 1 ). Since c (x 2 ) c (0), we have c (x 1 ) > c (0). Then, c(x) is strictly concave in an interval within [x 1, x 2 ] and at the same time strictly convex in an interval within [0, x 1 ]. It contradicts to Assumption 1. Thus, c (x) 0 throughout the domain and c(x) is convex. (2) Sub-additivity concavity: Let x > 0 and y > 0. By sub-additivity c(x + y) c(x) + c(y). With the zero fixed cost condition, we have 0. c(x + y) c(x) y c(y) c(0). y Let y 0, Thus, This means Let x 0, we obtain c(x + y) c(x) lim y 0 y c (x) c (0). c (x) c (0) x c(y) c(0) lim. y 0 y 0. c (0) = lim x 0 c (x) c (0) x Suppose that there exist x 2 > x 1 > 0 such that c (x 2) > c (x 1). Since c (x 2) c (0), we have c (x 1) < c (0). Then, c(x) is strictly convex in an interval within [x 1, x 2] and at the same time strictly concave in an interval within [0, x 1]. It contradicts to Assumption 1. Thus, c (x) 0 throughout the domain and c(x) is concave. 0. 7

9 3 Concluding Remark We usually assume that a cost function is convex throughout the domain or concave throughout the domain. Therefore, if there is no fixed cost, we can use interchangeably convexity and super-additivity of a cost function, and can use interchangeably concavity and subadditivity of a cost function. References Sen, D. and Stamatopoulos, G. (2016) Licensing under general demand and cost functions, European Journal of Operations Research, 253, pp Bruin, J.-C. and Hiai, F. (2015) Anti-norms on finite von Neumann algebras, Publications of the Research Institute for Mathematical Sciences, 51, pp Bruckner, A. M. and Ostrow, E. (1962) Some function classes related to the class of convex functions, Pacific Journal of Mathematics, 14, pp